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3 years ago

Kinetic energy in the form of moving waves of light or radiation

Chemistry

2 answers:

Vinvika [58]3 years ago

6 0

Answer:

radiation

Explanation:

adelina 88 [10]3 years ago

3 0

Answer:

neither

Explanation:

Kinetic Energy is the physical movement of objects.

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15. What are the outer electrons of an atom that have to ability to create<br> chemical bonds? *

suter [353]

Answer:

valence electron

Explanation:

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8 0

3 years ago

A book with a weight of 12 N rests on its back cover. If the back cover measures 0.21 m by 0.28 m, how much pressure does the bo

erik [133]

Answer:

option D = 204 Pa

Explanation:

Given data:

Weight of book = 12 N

Area of book = 0.21 m × 0.28 m = 0.0058 m²

Pressure of book = ?

Solution:

formula

pressure = force / area

P = f/ A

now we will put the values in formula,

P = 12 N/ 0.0058 m²

P= 204 Nm⁻² ( Nm⁻²= Pa)

P = 204 Pa

4 0

3 years ago

The Ba(OH)2 dissociates as Ba+2 + 2 OH-. H2SO4 dissociates as 2 H+ + SO4 -2.

suter [353]

Answer:

The balanced equations for those dissociations are:

Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻ (aq)

H₂SO₄ (aq) → 2H⁺(aq) + SO₄⁻²(aq)

Explanation:

As a strong base, the barium hidroxide gives OH⁻ to the solution

As a strong acid, the sulfuric acid gives H⁺ to the solution

Ba(OH)₂, is a strong base so the dissociation is complete.

H₂SO₄ is considerd a strong acid, but only the first deprotonation is strong.

The second proton that is released, has a weak dissociation.

H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺ (aq) + SO₄⁻² (aq) Ka

6 0

3 years ago

A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

Gnom [1K]

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.9440-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

0.0133 mole = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.171g$

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%$

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

7 0

4 years ago

You have competed an extraction procedure on a mixture of compounds known to contain Lauric acid ( it to get necessary physical

marin [14]

Answer:

3.5 mL

Explanation:

Let us represent lauric acid with the symbol LaCOOH and the anion as LaCOO-. The reaction between the anion and H2SO4 is;

LaCOO-(aq) + H2SO4(aq) -----> LaCOOH(aq) + HSO4-(aq)

Number of moles of LaCOO- = 2.75 g/200 g/mol = 0.014 moles

Since the mole ratio of the reaction is 1:1, then the amount of H2SO4 required is also 0.014 moles

Then;

n = CV

n = number of moles

C= concentration

V = volume

V = n/C

V = 0.014 moles/4.0 M

V= 3.5 * 10^-3 L

V = 3.5 mL

5 0

3 years ago